You are given a string time
in the form of hh:mm
, where some of the digits in the string are hidden (represented by ?
).
The valid times are those inclusively between 00:00
and 23:59
.
Return the latest valid time you can get from time
by replacing the hidden digits.
Example 1:
Input: time = "2?:?0"
Output: "23:50"
Explanation: The latest hour beginning with the digit '2' is 23 and the latest minute ending with the digit '0' is 50.
Example 2:
Input: time = "0?:3?"
Output: "09:39"
Example 3:
Input: time = "1?:22"
Output: "19:22"
Constraints:
time
is in the formathh:mm
.- It is guaranteed that you can produce a valid time from the given string.
给你一个字符串 time ,格式为 hh:mm(小时:分钟),其中某几位数字被隐藏(用 ? 表示)。有效的时间为 00:00 到 23:59 之间的所有时间,包括 00:00 和 23:59 。替换 time 中隐藏的数字,返回你可以得到的最晚有效时间。
- 简单题。根据题意,需要找到最晚的有效时间。枚举时间 4 个位置即可。如果第 3 个位置是 ?,那么它最晚时间是 5;如果第 4 个位置是 ?,那么它最晚时间是 9;如果第 2 个位置是 ?,那么它最晚时间是 9;如果第 1 个位置是 ?,根据第 2 个位置判断,如果第 2 个位置是大于 3 的数,那么第一个位置最晚时间是 1,如果第 2 个位置是小于 3 的数那么第一个位置最晚时间是 2 。按照上述规则即可还原最晚时间。
package leetcode
func maximumTime(time string) string {
timeb := []byte(time)
if timeb[3] == '?' {
timeb[3] = '5'
}
if timeb[4] == '?' {
timeb[4] = '9'
}
if timeb[0] == '?' {
if int(timeb[1]-'0') > 3 && int(timeb[1]-'0') < 10 {
timeb[0] = '1'
} else {
timeb[0] = '2'
}
}
if timeb[1] == '?' {
timeb[1] = '9'
}
if timeb[0] == '2' && timeb[1] == '9' {
timeb[1] = '3'
}
return string(timeb)
}